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Java-based LeetCode algorithm problem solutions, regularly updated
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49\. Group Anagrams
Medium
Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.
An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
**Example 1:**
**Input:** strs = ["eat","tea","tan","ate","nat","bat"]
**Output:** [["bat"],["nat","tan"],["ate","eat","tea"]]
**Example 2:**
**Input:** strs = [""]
**Output:** [[""]]
**Example 3:**
**Input:** strs = ["a"]
**Output:** [["a"]]
**Constraints:**
* 1 <= strs.length <= 104
* `0 <= strs[i].length <= 100`
* `strs[i]` consists of lowercase English letters.
To solve the "Group Anagrams" problem in Java with the Solution class, follow these steps:
1. Define a method `groupAnagrams` in the `Solution` class that takes an array of strings `strs` as input and returns a list of lists of strings.
2. Initialize an empty HashMap to store the groups of anagrams. The key will be the sorted string, and the value will be a list of strings.
3. Iterate through each string `str` in the input array `strs`.
4. Sort the characters of the current string `str` to create a key for the HashMap.
5. Check if the sorted string exists as a key in the HashMap:
- If it does, add the original string `str` to the corresponding list of strings.
- If it doesn't, create a new entry in the HashMap with the sorted string as the key and a new list containing `str` as the value.
6. After iterating through all strings, return the values of the HashMap as the result.
Here's the implementation of the `groupAnagrams` method in Java:
```java
import java.util.*;
class Solution {
public List> groupAnagrams(String[] strs) {
// Initialize a HashMap to store the groups of anagrams
Map> anagramGroups = new HashMap<>();
// Iterate through each string in the input array
for (String str : strs) {
// Sort the characters of the current string
char[] chars = str.toCharArray();
Arrays.sort(chars);
String sortedStr = new String(chars);
// Check if the sorted string exists as a key in the HashMap
if (anagramGroups.containsKey(sortedStr)) {
// If it does, add the original string to the corresponding list
anagramGroups.get(sortedStr).add(str);
} else {
// If it doesn't, create a new entry in the HashMap
List group = new ArrayList<>();
group.add(str);
anagramGroups.put(sortedStr, group);
}
}
// Return the values of the HashMap as the result
return new ArrayList<>(anagramGroups.values());
}
}
```
This implementation ensures that all anagrams are grouped together efficiently using a HashMap.
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